Упр.49.5 ГДЗ Мордкович 10-11 класс (Алгебра)

1. а) $$\int_{-1}^{0}\sqrt[3]{1-2x}\,dx$$; б) $$\int_{4}^{5}\frac{1}{(x-3)^3}\,dx$$; в) $$\int_{\frac{2}{3}}^{11}5\sqrt[5]{3x-1}\,dx$$; г) $$\int_{2}^{3}(5x-7)^{-\frac{2}{3}}\,dx$$.

а) $$\int_{-1}^{0}\sqrt[3]{1-2x}\,dx=\int_{-1}^{0}(1-2x)^{\frac13}\,dx$$

Сделаем замену: $$u=1-2x,\quad du=-2\,dx,\quad dx=-\frac12\,du.$$ Тогда

$$\int (1-2x)^{\frac13}\,dx=-\frac12\int u^{\frac13}\,du=-\frac12\cdot \frac{u^{\frac43}}{\frac43}=-\frac38u^{\frac43}.$$

Подставим пределы:

$$\int_{-1}^{0}\sqrt[3]{1-2x}\,dx=-\frac38(1-2x)^{\frac43}\Big|_{-1}^{0}=-\frac38\left(1^{\frac43}-3^{\frac43}\right).$$

Так как $$3^{\frac43}=3\sqrt[3]{3},$$ получаем

$$\int_{-1}^{0}\sqrt[3]{1-2x}\,dx=-\frac38\left(1-3\sqrt[3]{3}\right).$$

б) $$\int_{4}^{5}\frac{1}{(x-3)^3}\,dx=\int_{4}^{5}(x-3)^{-3}\,dx$$

Первообразная:

$$\int (x-3)^{-3}\,dx=\frac{(x-3)^{-2}}{-2}=-\frac{1}{2(x-3)^2}.$$

Тогда

$$\int_{4}^{5}\frac{1}{(x-3)^3}\,dx=-\frac{1}{2(x-3)^2}\Big|_{4}^{5}=-\frac{1}{2\cdot 2^2}+\frac{1}{2\cdot 1^2}=\frac38.$$

в) $$\int_{\frac23}^{11}5\sqrt[5]{3x-1}\,dx=\int_{\frac23}^{11}5(3x-1)^{\frac15}\,dx$$

Сделаем замену: $$u=3x-1,\quad du=3\,dx,\quad dx=\frac13\,du.$$ Тогда

$$\int 5(3x-1)^{\frac15}\,dx=\frac53\int u^{\frac15}\,du=\frac53\cdot \frac{u^{\frac65}}{\frac65}=\frac{25}{18}u^{\frac65}.$$

Подставим пределы:

$$\int_{\frac23}^{11}5\sqrt[5]{3x-1}\,dx=\frac{25}{18}(3x-1)^{\frac65}\Big|_{\frac23}^{11}=\frac{25}{18}\left(32^{\frac65}-1\right).$$

Так как $$32^{\frac65}=32\sqrt[5]{32}=64,$$ то

$$\int_{\frac23}^{11}5\sqrt[5]{3x-1}\,dx=\frac{25}{18}(64-1)=\frac{1575}{18}=87{,}5.$$

г) $$\int_{2}^{3}(5x-7)^{-\frac23}\,dx$$

Сделаем замену: $$u=5x-7,\quad du=5\,dx,\quad dx=\frac15\,du.$$ Тогда

$$\int (5x-7)^{-\frac23}\,dx=\frac15\int u^{-\frac23}\,du=\frac15\cdot \frac{u^{\frac13}}{\frac13}=\frac35u^{\frac13}.$$

Следовательно,

$$\int_{2}^{3}(5x-7)^{-\frac23}\,dx=\frac35(5x-7)^{\frac13}\Big|_{2}^{3}=\frac35\left(\sqrt[3]{8}-\sqrt[3]{3}\right).$$

Так как $$\sqrt[3]{8}=2,$$ получаем

$$\int_{2}^{3}(5x-7)^{-\frac23}\,dx=\frac35\left(2-\sqrt[3]{3}\right).$$

Ответ: а) $$-\frac38\left(1-3\sqrt[3]{3}\right);$$ б) $$\frac38;$$ в) $$87{,}5;$$ г) $$\frac35\left(2-\sqrt[3]{3}\right).$$